A 2×2 square is divided into four 1×1 squares. Each of the small squares is to be painted either green or red. In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square? There may be as few as zero or as many as four small green squares.

2

Step-by-step explanation:

Let's denote all the four square by 1, 2, 3, and 4 as shown in the figure.

Each of the small squares is to be painted either green or red

The total numbers of the possible cases are:

(1) The number of cases if there is no green color square = 1 (all red)

(2) The number of cases, if there is 1 green color square = 4 as either of 4 small squares can be green (1G,2G,3G,4G).

(3) The number of cases if there is 2 green color square = 2+4=6 as

(4) The number of cases, if there is 3 green color square = The number of cases, if there is 1 red color square= 4 as either of 4 small squares can be red (1R,2R,3R,4R).

(5) The number of cases if there is 4 green color square = 1 (all green)

So, the total numbers of possible cases = 1+4+6+4+1=16.

Now, possible positions for which no green square shares its top or right side with any red square are the diagonal positions as mentioned in point (3), i.e (1G,3G), (2G,4G)

So, the total numbers of favorable cases = 2

Hence the total number of different ways for painting each of the small squares so that no green square shares its top or right side with any red square is 2.

the answer is c

hope this !

the least value out of the data shown is 5, maybe if you post a picture with it i could a little more.

step-by-step explanation: